Respuesta :
The third string is the longest.
a = first string
b = second string
c = third string
The first column, closest to the left, is the equations derived from the problem. Because the last two equations all have "b" in common, replace "a" and "c" in the first equation with what they equal to. (a+b+c=36 becomes [(2b) + b + (2b+1) = 36]) Then, solve for "b". Once you have found "b", plug it in the equation for the first and third string; compare.
(The reason why I only said the third string is the longest and not the length specifically, is so you can figure out the final answer completely on your own. )
a = first string
b = second string
c = third string
The first column, closest to the left, is the equations derived from the problem. Because the last two equations all have "b" in common, replace "a" and "c" in the first equation with what they equal to. (a+b+c=36 becomes [(2b) + b + (2b+1) = 36]) Then, solve for "b". Once you have found "b", plug it in the equation for the first and third string; compare.
(The reason why I only said the third string is the longest and not the length specifically, is so you can figure out the final answer completely on your own. )

Answer:
The longest piece is 21 feet.
Step-by-step explanation:
Let the three pieces be x,y and z.
The first piece of ribbon is half as long as the second piece of ribbon.
[tex]x= \frac{y}{2}[/tex] ....(1)
The third piece is 1 foot longer than twice the length of the second piece of ribbon.
[tex]z=2y+1[/tex] .........(2)
And together all pieces measure 36 feet.
[tex]x+y+z=36[/tex] ......(3)
Substituting values of x and z in (3)
[tex]\frac{y}{2}+ y+2y+1=36[/tex]
Solving this we get;
[tex]\frac{y+2y+4y+2}{2}=36[/tex]
=> [tex]\frac{7y+2}{2}=36[/tex]
=> [tex]7y+2=72[/tex]
=> [tex]7y=70[/tex]
y = 10 feet
As [tex]x= \frac{y}{2}[/tex]
[tex]x= \frac{10}{2}[/tex]
x = 5 feet
And also [tex]x+y+z=36[/tex]
=> [tex]10+5+z=36[/tex]
=> [tex]15+z=36[/tex]
=> [tex]z=36-15[/tex]
z = 21 feet
Hence, the longest piece is 21 feet.